Show that the spaces \((S^1 \times \mathbb{CP}^\infty) / (S^1 \times \{x_0\})\) and \(S^3 \times \mathbb{CP}^\infty\) have isomorphic cohomology rings with \(\mathbb{Z}\) or any other coefficients. [An exercise for §4.L is to show these two spaces are not homotopy equivalent.]
Solution
Let \(Y=\mathbb{CP}^\infty\) with basepoint \(y_0\), and set \[ Q=(S^1\times Y)/(S^1\times\{y_0\}). \] Reduced cohomology of \(Q\) is relative cohomology: \[ \widetilde H^*(Q;R)\cong H^*(S^1\times Y,S^1\times\{y_0\};R) \] for any coefficient ring \(R\).
Write \(u\in H^2(Y;R)\) for the standard generator and \(s\in H^1(S^1;R)\). The relative Künneth theorem gives \[ H^*(S^1\times Y,S^1\times\{y_0\};R) \cong H^*(S^1;R)\otimes\widetilde H^*(Y;R). \] Thus, additively, the positive-degree classes are \[ u^k\quad(k\ge1),\qquad s u^k\quad(k\ge1). \] The relative cross product is compatible with cup products, so \[ u^i u^j=u^{i+j},\qquad (su^i)u^j=su^{i+j},\qquad (su^i)(su^j)=0 \] because \(s^2=0\). If \[ v=su\in H^3(Q;R), \] then every odd class is \(vu^{k-1}\) and \(v^2=0\). Hence \[ H^*(Q;R)\cong R[u,v]/(v^2),\qquad |u|=2,\ |v|=3. \]
On the other hand \[ H^*(S^3\times\mathbb{CP}^\infty;R) \cong H^*(S^3;R)\otimes_R R[u] \cong R[u,v]/(v^2), \] with the same degrees. Therefore the two spaces have isomorphic cohomology rings with \(\mathbb Z\) coefficients and, by the same calculation, with arbitrary coefficients.